Higher. Straight Lines 1

Transcription

1 hsn uknet Higher Matheatics UNIT 1 OUTOME 1 Straight Lines ontents Straight Lines 1 1 The Distance etween Points 1 The Midpoint Forula 3 3 Gradients 4 4 ollinearit 6 5 Gradients of Perpendicular Lines 7 6 The Equation of a Straight Line 8 7 Medians 11 8 ltitudes 1 9 Perpendicular isectors Intersection of Lines oncurrenc 17 HSN1100 This docuent was produced speciall for the HSNuknet website, and we require that an copies or derivative works attribute the work to Higher Still s For ore details about the copright on these notes, please see

2 OUTOME 1 Straight Lines 1 The Distance etween Points Points on Horizontal or Vertical Lines It is relativel straightforward to work out the distance between two points which lie on a line parallel to the - or -ais O O (, ) (, ) 1 1 d EXMPLE d (, ) (, ) 1 1 In the diagra to the left, the points ( 1, 1) and (, ) lie on a line parallel to the -ais, ie 1 The distance between the points is sipl the difference in the -coordinates, ie d 1 where > 1 In the diagra to the left, the points ( 1, 1) and (, ) lie on a line parallel to the -ais, ie 1 The distance between the points is sipl the difference in the -coordinates, ie d 1 where > 1 1 alculate the distance between the points ( 7, 3) and ( 16, 3) Since both -coordinates are 3, the distance is the difference in the - coordinates: d 16 ( 7) units Page 1 HSN1100

3 The Distance Forula The distance forula gives us a ethod for working out the length of the straight line between an two points It is based on Pthagoras s Theore (, ) The 1 and d 1 1 coe fro the ethod above ( 1, 1 ) 1 O The distance d between the points (, ) and (, ) EXMPLES is the point (, 4) The length is 1 1 ( ) ( ) 1 1 is d + units and ( 3,1 ) alculate the length of the line ( ) + ( ) 1 1 ( 3 ( ) ) + ( 1 4) 5 + ( 3) units 15 3 alculate the distance between the points ( ) The distance is ( ) + ( ) 1 1 ( 1 15 ) ( 1 ) ( ) ( ) ( ) ( ) units 4 1, 4 and ( 1, 1) You need to becoe confident working with fractions and surds so practise! Page HSN1100

4 The Midpoint Forula The point half-wa between two points is called their idpoint It is calculated as follows The idpoint of (, ) and (, ) 1 1 is + +, 1 1 It a be helpful to think of the idpoint as the average of two points EXMPLES 1 alculate the idpoint of the points ( 1, 4) The idpoint is + +, 1 1 ( 4, ) ( ) , and ( 7, 8 ) In the diagra below, ( 9, ) lies on the circuference of the circle with centre ( 17,1 ), and the line is a diaeter of the circle Find the coordinates of Sipl writing The idpoint is (4, ) would be acceptable in an ea Since is the centre of the circle and is a diaeter, is the idpoint of Using the idpoint forula, we have: ( 17,1 ), where is the point (, ) coparing - and -coordinates, we have: and So is the point ( 5, 6 ) 6 Page 3 HSN1100

5 3 Gradients onsider a straight line passing through the points (, ) and (, ) 1 1 The gradient of the line through (, ) and (, ) change in vertical height 1 1 is 1 1 change in horizontal distance 1 Opposite 1 lso, since tanθ djacent O (, ) 1 we obtain: tanθ where θ is the angle between the line and the positive direction of the -ais θ 1 1 θ (, ) 1 1 for : θ is the Greek letter theta It is often used to stand for an angle positive direction s a result of the above definitions: lines with positive gradients slope up, fro left to right; lines with negative gradients slope down, fro left to right; lines parallel to the -ais have a gradient of zero; lines parallel to the -ais have an undefined gradient We a also use the fact that: Two distinct lines are said to be parallel when the have the sae gradient (or when both lines are vertical) Page 4 HSN1100

6 EXMPLES 1 alculate the gradient of the straight line shown in the diagra below tanθ tan3 O (to dp) Find the angle that the line joining P(, ) and Q ( 1, 7 ) akes with the positive direction of the -ais The line has gradient nd so tanθ tanθ 3 1 θ tan ( 3) (to dp) 3 Find the size of angle θ shown in the diagra below θ 5 O We need to be careful because the θ in the question is not the θ in tanθ So we work out the angle a and use this to find θ: a tan 1 ( ) 1 tan ( 5) So θ (to dp) O θ a 5 Page 5 HSN1100

7 4 ollinearit Points which lie on the sae straight line are said to be collinear To test if three points, and are collinear we can: 1 Work out Work out (or ) 3 If the gradients fro 1 and are the sae then, and are collinear so, and are collinear If the gradients are different then the points are not collinear so, and are not collinear This test for collinearit can onl be used in two diensions EXMPLES 1 Show that the points P( 6, 1), Q ( 0, ) and R ( 8, 6 ) are collinear ( 1) PQ 0 ( 6) QR Since PQ and Q is a coon point, P, Q and R are collinear QR The points ( 1, 1), ( 1,k ) and ( 5,7 ) are collinear Find the value of k Since the points are collinear : k ( 1) 7 ( 1) k k + 1 ( ) k 5 Page 6 HSN1100

8 5 Gradients of Perpendicular Lines Two lines at right-angles to each other are said to be perpendicular If perpendicular lines have gradients and then onversel, if 1 1 then the lines are perpendicular The siple rule is: if ou know the gradient of one of the lines, then the gradient of the other is calculated b inverting the gradient (ie flipping the fraction) and changing the sign For eaple: 3 if then 3 that this rule cannot be used if the line is parallel to the - or -ais If a line is parallel to the -ais ( 0), then the perpendicular line is parallel to the -ais it has an undefined gradient If a line is parallel to the -ais then the perpendicular line is parallel to the -ais it has a gradient of zero EXMPLES 1 Given that T is the point ( 1, ) line perpendicular to ST ST 5 ( ) 4 1 So 5 7 since ST and S is ( 4, 5), find the gradient of a Triangle MOP has vertices M( 3, 9), O( 0, 0 ) and P( 1, 4 ) Show that the triangle is right-angled Sketch: M( 3, 9) O( 0, 0) P( 1, 4) OM MP OP Since OM OP 1, OM is perpendicular to OP which eans MOP is right-angled at O The converse of Pthagoras s Theore could also be used here: Page 7 HSN1100

9 d d OP ( 3) OM d ( 1 ( 3) ) + ( 4 9) MP 15 + ( 5) 50 Since d + d d, triangle MOP is right-angled at O OP OM MP 6 The Equation of a Straight Line To work out the equation of a straight line, we need to know two things: the gradient of the line, and a point which lies on the line The straight line through the point ( a, b) b ( a) with gradient has the equation Notice that if we have a point ( 0, c ) the -ais intercept then the equation becoes + c You should alread be failiar with this for It is good practice to rearrange the equation of a straight line into the for a + b + c 0 where a is positive This is known as the general for of the equation of a straight line Lines Parallel to es c c O If a line is parallel to the -ais (ie 0 ), its equation is c O k k If a line is parallel to the -ais (ie is undefined), its equation is k Page 8 HSN1100

10 EXMPLES 1 Find the equation of the line with gradient 1 3 ( 3, 4) b ( a) ( ) ( ) ( 3) passing through the point Find the equation of the line passing through ( 3, ) and (,1) To work out the equation, we ust first find the gradient of the line : ( ) 5 3 Now we have a gradient, and can use this with one of the given points: b ( a) 1 ( ) ( ) using 3, and Find the equation of the line passing through ( 3 ) and ( 3 ) 5, 4 The gradient is undefined since the -coordinates are equal So the equation of the line is 3 5 It is usuall easier to ultipl out the fraction before epanding the brackets 5, 5 Page 9 HSN1100

11 Etracting the Gradient You should alread be failiar with the following fact The line with equation + c has gradient It is iportant to reeber that ou ust rearrange the equation of a straight line into this for before etracting the gradient EXMPLES 4 Find the gradient of the line with equation We have to rearrange the equation: So the gradient is 3 5 The line through points ( 3, 3) and has equation Find the equation of the line through which is perpendicular to First, find the gradient of : So 5 and 1 5 Therefore the equation is: + 3 ( 3 ) using ( 3, 3 ) and ( 3) Page 10 HSN1100

12 7 Medians edian of a triangle is a line through a verte and the idpoint of the opposite side M is a edian of The standard process for finding the equation of a edian is shown below EXMPLE M Triangle has vertices ( 4, 9) ( 10, ) and ( 4, 4), Find the equation of the edian fro Start with a sketch: M Step 1 alculate the idpoint of the relevant line Step alculate the gradient of the line between the idpoint and the opposite verte Step 3 Find the equation using this gradient and either of the two points used in Step 10, and ( 4, 4) ( 4) Using ( ) M, 14, ( 7, 1 ) Using ( 4, 9) M Using ( 4, 9) : and M( 7, 1) : ( 9) and M : b ( a) 8 ( ) ( 3) Page 11 HSN1100

13 8 ltitudes n altitude of a triangle is a line through a verte, perpendicular to the opposite side D is an altitude of D The standard process for finding the equation of an altitude is shown below EXMPLE Triangle has vertices ( 3, 5), 4, 3 and ( 7, ) ( ) Find the equation of the altitude fro Start with a sketch: D Step 1 alculate the gradient of the side which is perpendicular to the altitude Step alculate the gradient of the altitude using 1 Step 3 Find the equation using this gradient and the point that the altitude passes through 4, 3 and ( 7, ) : Using ( ) Using D 1: D 11 Using ( 3, 5) and D 11: b ( a) ( 3) Page 1 HSN1100

14 9 Perpendicular isectors perpendicular bisector is a line which cuts through the idpoint of a line segent at right-angles D In both cases, D is the perpendicular bisector of The standard process for finding the equation of a perpendicular bisector is shown below D EXMPLE is the point (,1) and is the point ( 4, 7 ) Find the equation of the perpendicular bisector of Start with a sketch: Step 1 alculate the idpoint of the line segent being bisected Step alculate the gradient of the line used in Step 1, then find the gradient of its perpendicular bisector using 1 Step 3 Find the equation of the perpendicular bisector using the point fro Step 1 and the gradient fro Step Using (,1) and ( 4, 7 ) : Midpoint + +, 1, 4 Using (,1) ( ) 4, 7 : and ( ) ( ) since 1 Using ( 1, 4 ) and 1: b ( a) 4 ( 1) Page 13 HSN1100

15 10 Intersection of Lines Man probles involve lines which intersect (cross each other) Once we have equations for the lines, the proble is to find values for and which satisf both equations, ie solve siultaneous equations There are three different techniques and, depending on the for of the equations, one a be ore efficient than the others We will deonstrate these techniques b finding the point of intersection of the lines with equations and 3 Eliination This should be a failiar ethod, and can be used in all cases : 18 9 Put 9 into : So the lines intersect at the point ( 1, 9 ) Equating This ethod can be used when both equations have a coon - or - coefficient In this case, both equations have an -coefficient of 1 Make the subject of both equations: Equate: Substitute 9 into: So the lines intersect at the point ( 1, 9 ) Page 14 HSN1100

16 Substitution This ethod can be used when one equation has an - or -coefficient of 1 (ie just an or with no ultiplier) Substitute 3 into: ( 3) So the lines intersect at the point ( 1, 9 ) EXMPLE Substitute 1 into: Find the point of intersection of the lines and Eliinate : : Put 3 into 1: So the point of intersection is ( 3,5) Triangle PQR has vertices P( 8,3 ), Q ( 1,6 ) and R (, 3) Q T M P S (a) Find the equation of altitude QS (b) Find the equation of edian RT (c) Hence find the coordinates of M R Page 15 HSN1100

17 (a) Find the gradient of PR: PR 3 ( 3) So the gradient of QS is QS 1, since the PR QS 1 Find the equation of QS using Q ( 1,6 ) and QS 1: 6 1( + 1) (b) Find the coordinates of T, the idpoint of PQ: T, 7 9 (, ) Find the gradient of RT using R (, 3) and 7 9 ( ) T, : This is the standard ethod for finding the equation of an altitude This is the standard ethod for finding the equation of a edian RT 9 15 ( 3) Find the equation of RT using R (, 3) and RT 5: + 3 5( ) (c) Now solve the equations siultaneousl to find M Eliinate : : Put 3 into 1: So the point of intersection is M( 3, ) n of the three techniques could have been used here Page 16 HSN1100

18 11 oncurrenc n nuber of lines are said to be concurrent if there is a point through which the all pass So in the previous section, b finding a point of intersection of two lines we showed that the two lines were concurrent For three lines to be concurrent, the ust all pass through a single point The three lines are concurrent The three lines are not concurrent surprising fact is that the following lines in a triangle are concurrent The three edians of a triangle are concurrent The three altitudes of a triangle are concurrent The three perpendicular bisectors in a triangle are concurrent The three angle bisectors of a triangle are concurrent Page 17 HSN1100